Problem
ALG-B3-M08-P004 Affine Involution
#4
★★☆☆☆ Level 2 of 5
Find all functions \(f(x)=ax+b\) such that \(f(f(x))=x\).
Compute \(f(f(x))\).
We have \(f(f(x))=a(ax+b)+b=a^2x+b(a+1)\). Thus \(a^2=1\) and \(b(a+1)=0\). If \(a=1\), then \(b=0\), giving \(f(x)=x\). If \(a=-1\), then \(b\) is arbitrary, giving \(f(x)=-x+b\).
Shows that an involution need not be the identity.